Question

Evaluate the derivative of the following functions at the given point. $$f(t)=\frac{1}{t+1} ; a=1$$

Short answer

Answer: \(-\frac{1}{4}\)

Step-by-step solution

Identify u(t) and v(t)

In this problem, we have: $$u(t) = 1$$ $$v(t) = t+1$$

Calculate the derivatives of u(t) and v(t)

Now, compute the derivatives of \(u(t)\) and \(v(t)\): $$u'(t) = \frac{d}{dt}1 = 0$$ $$v'(t) = \frac{d}{dt}(t+1) = 1$$

Apply the quotient rule for differentiation

Using the quotient rule, we can calculate the derivative of the given function: $$f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{v^2(t)}$$ Substitute the derivatives and functions we found earlier: $$f'(t) = \frac{(0)(t+1) - (1)(1)}{(t+1)^2}$$ Simplify the expression: $$f'(t) = \frac{-1}{(t+1)^2}$$

Evaluate the derivative at the given point

The last step is to evaluate \(f'(t)\) at \(t=1\): $$f'(1) = \frac{-1}{(1+1)^2} = \frac{-1}{4}$$ So, the derivative of the function \(f(t)=\frac{1}{t+1}\) at the point \(t=1\) is \(-\frac{1}{4}\).

Key concepts

Quotient Rule for Differentiation

Understanding how to work with fractions in calculus is essential, especially when it comes to the differentiation of functions that are presented as a quotient, like \( f(t) = \frac{1}{t+1} \). The quotient rule is a method used for finding the derivative of a function that is the ratio of two differentiable functions.

The quotient rule states that if you have a function \( g(t) = \frac{u(t)}{v(t)} \) where both \( u(t) \) and \( v(t) \) are differentiable, the derivative \( g'(t) \) is calculated as follows:\[ g'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{v^2(t)} \]Breaking it down, \( u'(t) \) and \( v'(t) \) are the derivatives of \( u(t) \) and \( v(t) \) respectively. Then you multiply \( u'(t) \) by \( v(t) \) and subtract the product of \( u(t) \) and \( v'(t) \) from it. Finally, divide the result by the square of \( v(t) \). Precisely calculating these derivatives and correctly applying the quotient rule is crucial in solving calculus problems accurately.

Evaluating Derivatives at a Point

After determining the derivative of a function, the next step is often to 'plug in' a specific value for the variable to evaluate the derivative at that point. This process is used to find the rate of change or the slope of the tangent line to the curve at a particular value.

When you evaluate the derivative of a function like \( f'(t) \) at \( t = a \), you are essentially replacing every occurrence of \( t \) with \( a \) in your derivative equation.In the given exercise, once we have the derived function \( f'(t) = \frac{-1}{(t+1)^2} \) using the quotient rule, to evaluate it at \( t = 1 \) simply requires us to substitute \( 1 \) for \( t \) in our expression:\[ f'(1) = \frac{-1}{(1+1)^2} = \frac{-1}{4} \]This action yields the slope of the curve at \( t = 1 \) which, in this case, is \( -\frac{1}{4} \) indicating a negative slope. Evaluating derivatives at particular points is essential in many applications in physics and engineering, such as finding velocities, accelerations, and optimizing functions.

Calculus Problem Solving

Tackling calculus problems requires a combination of conceptual understanding and procedural skills. The problem from the textbook demonstrates a step-by-step approach, which is integral to calculus problem solving. Start by carefully identifying the components of the function and use the appropriate rules for differentiation, such as the quotient rule in this example.

Effective problem solving in calculus also involves:

• Understanding the definitions and theorems involved, like the quotient rule.
• Correctly applying these rules to the function you have.
• Simplifying your results to make evaluation easier.
• Checking your work to avoid calculation errors.

Finally, practice is key. The more problems you solve, the more familiar you’ll become with the types of functions and the various differentiation rules. This familiarity will help you identify the best approach to each new problem you encounter, leading to more efficient and accurate solutions. When students follow these steps and truly comprehend each one, calculus becomes much less daunting, allowing for the mastery of even complex problems.